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I know that there is a following version of the Weierstrass approximation theorem for functions on $[a,b]$. For every $f:[a,b] \rightarrow \mathbb{R}$ there is a sequence $(P_n)$ of polynomials such that $P_n^{(i)}(x) \rightrightarrows f^{(i)}(x) $ as $n \rightarrow \infty$, for $i=0,1,...,k$ .

Is there analogue of this theorem for periodic functions?

Assume that a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is $2\pi$-periodic of class $C^k$ and $f^{(i)}(2m\pi)=0$ for $i=1,...,k$; $m \in \mathbb{Z}$. Is it true that there exists a sequence $(P_n)$ of trigonometric polynomials such that $P_n^{(i)}(x) \rightrightarrows f^{(i)}(x) $ as $n \rightarrow \infty$, for $i=0,1,...,k$ ?

Thanks.

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1 Answer 1

up vote 2 down vote accepted

I had to revise the answer somewhat :-). Probably it is in fact not that easy to use the Stone Weierstrass Theorem for this, but the approach using Fourier Series works out fine.

The $N$-th Fourier polynomial of a continuous periodic function (on $[-\pi,\pi]$, say) is given by $$ S_N(f)(x) := \sum_{k=0}^N c_k(f) e^{ikx}$$ where $$ c_k(f) := \frac{1}{2\pi} \int_{-\pi}^{\pi}f(y) e^{-iky}dy$$

The Fourier Series of $f$ corresponds to $N=\infty.$ If $f$ is differentiable and $f^'$ integrable such the fundamental theorem of calculus holds, it is easy to see (by partial integration) that $$\frac{d}{dx}S_N(f)(x)= \sum_{k=0}^Nc_k(f') e^{-ikx}$$ that is, the derivative of the $N$-th Fourier polynomial is the $N$-th Fourier polynomial of the derivative. (All this can be found in Rudin's "Principles of Mathematical Analysis").

While, in general, it will not be true that the Fourier series of a $C^k$ function $f$ converges together with the first $k$ derivatives uniformly to $f$ and it's derivatives, this will be true if $f$ is, e.g., piecewise $C^{k+1}$. This is shown in most textbooks on analysis for piecewise differentiable $f$ (e.g. in Rudin's book mentioned above), and because of the aforementioned relation between fourier coefficients and taking the derivatives this result may be applied to $f^{(k)}$ to prove the same statement for piecewise $C^{k+1}$ functions.

But piecewise $C^{k+1}$ functions are dense in $C^k$. Consequently, trigonometric polynomials are, too.

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Very thanks for nice answer. –  Richard Jan 2 '12 at 18:49

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